# Equivalence of viscosity and weak solutions for the Poisson equation

Suppose $$\Omega$$ is a bounded smooth domain in $$\mathbb{R}^d$$.

How does one prove that weak solutions are viscosity solutions and vice versa for the problem $$\begin{cases} -\Delta u = f(x) & \text{ in } \Omega\\ u=g & \text{ on } \partial \Omega \end{cases}$$ under suitable assumptions (to be determined) on $$f$$ and $$g$$?

The general proof is in this paper, but I'm looking for a simple proof in the case of this toy problem. I wouldn't want to rely on deep general results on elliptic regularity.

• I believe you do need at least some a priori regularity results for that. For example, if there existed a discontinuous weak solution $u$, by definition it would fail to be a viscosity solution. However, I am not an expert here. Apr 14, 2019 at 9:46
• Why do you want to do this without elliptic regularity? If you use classical solvability of your problem, with $f$ and $g$ sufficiently smooth, then there is a very simple proof that the viscosity solution is unique and agrees with the classical solution. Without using elliptic regularity, you may need the proofs in the Ishii paper you referenced, which use inf and sup-convolutions.
– Jeff
Apr 15, 2019 at 1:20
• @Jeff Can the proof in Ishii paper be done in a simplified way if the operator is the Laplacian?
– user123456
Apr 15, 2019 at 12:11
• It doesn't simplify much. The ideas in the paper are standard tools in viscosity solution theory (inf and sup convolutions). Probably it is a good idea to learn the tools and understand Ishii's proof.
– Jeff
Apr 15, 2019 at 20:04
• BTW, if you allow classical solvability for $f$ and $g$ smooth, the proof goes by comparing your viscosity solution against strict super/sub solutions. The comparison principle between a viscosity solution, and a smooth strict sub/super solution is trivial from the definition of viscosity solutions.
– Jeff
Apr 15, 2019 at 20:07